ABC is any triangle and O is any point in the plane of the triangle. AO, BO, CO meet the sides BC, CA, AB in D, E, F respectively. 

Find $\frac{OD}{AD}+\frac{OE}{BE}+\frac{OF}{CF}$

Take O as the origin. 
Let $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ be the position vectors of the vertices A, B, C of the triangle. 
Since $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are coplanar there must exist a relation of the form 
$x\overrightarrow{a}+y\overrightarrow{b}+z\overrightarrow{c}=\overrightarrow{0}$,  (x, y, z$\in R$ , not  all zero)     . . . (i) 
$y\overrightarrow{b}+z\overrightarrow{c}=-x\overrightarrow{a}$
 $\Rightarrow \frac{y\overrightarrow{b}+z\overrightarrow{c}}{y+z}=-\frac{x}{y+z}\overrightarrow{a}$
Now $\frac{y\overrightarrow{b}+z\overrightarrow{c}}{y+z}$ is a point on the line BC. 

Equation of $\overrightarrow{OA}$ is $\overrightarrow{r}=-\overrightarrow{ta}$
Thus $-\frac{x}{y+a}\overrightarrow{a}$ is a point of AD.
From (ii), $\overrightarrow{OD}=\frac{-x}{y+z}\overrightarrow{a}$ 
$\frac{OD}{AD}=\frac{\left|\overrightarrow{OD}\right|}{\left|\overrightarrow{AD}\right|}=\left(-\frac{{\displaystyle x}}{{\displaystyle y+z}}\right)\times \left(-\frac{{\displaystyle y+z}}{{\displaystyle x+y+z}}\right)$$=\frac{x}{x+y+z}$

Similarly, $\frac{OE}{BE}=\frac{y}{x+y+z}$ and $\frac{OF}{CF}=\frac{z}{x+y+z}$ 
Adding, $\frac{OD}{AD}+\frac{OE}{BE}+\frac{OE}{CF}=\frac{x+y+z}{x+y+z}=1$